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   The Number Pyramid
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   Author  Topic: The Number Pyramid  (Read 6909 times)
Naumz
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The Number Pyramid  
« on: Oct 3rd, 2006, 6:15pm »
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A Number Pyramid is composed of the 10 different numbers 0-9 with a top row of 1 number resting on a second row of 2 sitting on a third row of 3 supported by a bottom row of 4. For example, a Number Pyramid could be:
 
 
   0
  1 2
 3 4 5  
6 7 8 9

 
Given the clues below, can you determine the composition of the Number Pyramid?
a) The number at the top of Number Pyramid minus the leftmost number in row 2 equals 5.
b) Both the four leftmost numbers in each row and the three numbers in row 3 sum to 14.
c) In the bottom row of the pyramid, the second number from the left minus the leftmost number equals 5.
d) The rightmost number in row 3 minus the rightmost number in row 4 equals 3.
e) The four numbers in the bottom row sum to 17.
« Last Edit: Oct 4th, 2006, 6:15pm by Icarus » IP Logged
Naumz
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Re: The Number Pyramid  
« Reply #1 on: Oct 3rd, 2006, 6:16pm »
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The last row is shifted a bit to the right, not that it matters, of course. But the formatting looks weird...loll  Grin
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towr
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Re: The Number Pyramid  
« Reply #2 on: Oct 4th, 2006, 12:19am »
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Seems a simple matter of solving a system of equations.
with a further constraint all numbers 0-9 are used.
 
a-b=5
a+b+d+g=14 & d+e+f=14
h-g=5
f-j=3
g+h+i+j=17
 
A computer would get it in a second, of course..
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Naumz
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Re: The Number Pyramid  
« Reply #3 on: Oct 4th, 2006, 11:22am »
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I did solve it by a computer. But i thought there would be a smipler way to do by ourselves without solving the equations completely...guess i was wrong. I actually got this as part of a project.
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Re: The Number Pyramid  
« Reply #4 on: Oct 4th, 2006, 12:59pm »
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It's probably not impossible to do it by hand. It's just that I'm lazy, and would rather use a simple solver than think twice  Roll Eyes
The standard way to do it by hand would probably to make a 10 by 10 table (variables vertically, digits horizontally) and start eliminating ranges that don't work.
e.g. a-b = 5 gives 5 =< a =< 9 and 0 =< b =< 4
It's tricky for multiple variables in an equation though.
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Re: The Number Pyramid  
« Reply #5 on: Oct 4th, 2006, 6:18pm »
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on Oct 3rd, 2006, 6:16pm, Naumz wrote:
The last row is shifted a bit to the right, not that it matters, of course. But the formatting looks weird...loll  Grin

 
I reformatted it for you. YaBB does a lot of weird things to formatted text. In particular, it doesn't allow more than about 5 or 6 spaces together. When it sees that many it always replaces them with a single space. Thats why I removed the "indent". (I also used a constant width font, so that spaces are just as wide as the characters.)
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Re: The Number Pyramid  
« Reply #6 on: Oct 4th, 2006, 6:32pm »
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Since you only have 6 equations, the best you can do is solve for 6 variables in terms of the other 4. Since one variable (c) is not involved in any of the equations, we have 3 variables that determine the other 6, and one left to take on whatever value is left over.
 
While the equations place some limitations on what values the 3 "independent" variables take on (for instance, g has to be >= 3), for the most part, at this stage, you are reduced to using brute force to finish solving the puzzle by simply trying out triplets of values and see which ones result in all values from 0-9 except one in the 3 independent and 6 dependent variables (the remaining value is the value of c).
 
And since brute force is what is called for at this stage, what better way to apply it than to use a computer?
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Re: The Number Pyramid  
« Reply #7 on: Oct 4th, 2006, 6:48pm »
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In addition to the equations towr gave, there is also 45=a+b+c+d+e+f+g+h+i+j. From this subtract some of towr's equations to get c=d+g. Substitute to remove h, a, d, and g from the equations:
(1)  2*b+c=9
(2)  c-g+e+j=11
(3)  2*g+i+j=12
 
From (1) and a-b=5, b can only be 0, 1, 2 and the corresponding values of c are 9, 7, or 5 (b can't be 3 or c would be 3 too, and b can't be 4 because c would be 1 and d or g would also have to be 1). Since all letters must have different values, c=d+g,  g<5 (because h=g+5) there are not many possibilities (keeping in mind that h=g+5 must also have a different value):
 
b=0, c=9, (d=8,g=1 or d=6,g=3)
b=1, c=7, (d=4,g=3 or d=3,g=4)
b=2, c=5, (d=1,g=4 or d=4,g=1)
 
Eliminate j from (2) and (3) to get: 1+c-3*g=i-e. Plug in the possible values of c and g into this equation and try to find acceptable values of i and e that do not duplicate any of the other numbers (make sure i and e are also different from h=g+5 and a=b+5). The remaining possibilites are now:
 
b=0, c=9, d=6, g=3, i=2, e=1
b=2, c=5, d=1, g=4, i=0, e=6
b=2, c=5, d=4, g=1, i=3, e=0
 
Only the row with b=0 permits values of j and f with f=j+3 which do not duplicate other numbers (j=4, f=7), and the only acceptable values are
 
a=5, b=0, c=9, d=6, e=1, f=7, g=3, h=8, i=2, j=4
 
Hmmm, that seemed much easier before I tried to type out a full explanation.
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